By Uggappakodi Narayan Bhat

**Read or Download A Study of the Queueing Systems M/G/1 and GI/M/1 PDF**

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**Extra info for A Study of the Queueing Systems M/G/1 and GI/M/1**

**Sample text**

78) we have the theorem. 85). In this and the following theorem, we shall not be simplifying the expressions derived by probabilistic arguments. easier to simplify in special cases. 6. 6 Waiting time W(t): An approach through Q(t). We define the waiting time process W(t) as follows. Suppose there is a mechanism (say, an inspector) that observes the system as to how long a new customer at time enters his service. t, Clearly, at any time if any, will have to wait till he t, with queue discipline 'first come- forst served' this would be the time required for all the customers in the system to complete their service.

M (1. 73) can be further simplified as ~ n~j ('t)k b(k) ' oo _A l l L b dG (t) of00 e-At n+j n=O k=O k! 74) j > 0, Thus we get, for = A(l-p) p~ J - foo 0 e-At n+j oo L L b(k~ (At) n=O k=O n+J k! k C (t)dt] n (1. 60) by noting that 0f oo e -At n+~-l (k+l) (At)k [Bn(t)- Bn+l(t))dt k! k~O bn+j = T=Of oo dB - <=Of oo n (t) t=T fooe-At n+j-1 k~O oo dBn+l (t) t=Tf e -At n+t"-1 k=O and = n+j-l L k=O b (k+l) k +" n J Le 0 b(k+l)(A~)k -AT (AT)r 1 r. n+j k. dt b (k+l) (At) k! n+j k dt (1. 76) (1. 77) is obtained by repeated integration by parts of the left hand side.

We do this by studying a related process Y*(t) = u + t - X(t). 91) (u > 0) The stochastic process Y*(t): Let {X(t) , 0 ~ t ~ T} be a separable stochastic process with non-negative increments, increasing only in jumps and continuous on the right. Let N(t) be the number of such increases in X(t). Further, let these jumps occur in a Poisson process and the magnitude of the jumps have a distribution distribution of X(t) and dB(x) (0 < x < oo) N(t) is given by such that, the joint 36 = Pr{X(t) K (x,t) n ~ x , N(t) n e-At (>,;) B (x) n.